Recent content by Cole.

Ahh but it does! Silly mistake from my part. Thank you so much it has been a real journey :) Regards Cole.

Yes I am referring to the last term: \dfrac{1}{2}m( \dot{r}+r \dot{\theta} )^{2}=\dfrac{1}{2}m(\dot{r}^{2}+2r\dot{r} \dot{\theta}+r^{2}\theta ^{2}) What i think you mean in the above is to say: T_{b}=\dfrac{1}{2}I_{b}\dot{\theta}+\dfrac{1}{2}m \dot{r} ^{2}+\dfrac{1}{2}mr^{2} \dot{\theta}^{2} but...

Everything appears correct except for the velocity factor. When I expand it I get: \dfrac{1}{2}m\ddot{r}^{2}+m\dot{r}r\dot{\theta}+ \dfrac{1}{2} mr^{2}\dot{\theta}^{2} and when I use the part of the Legrender equation relevant for the energy I get...

Great i figured it out! However I have run into a new problem. Suppose the rod has a bead of mass m attached at the pivot moving down is it rotates. I can't figure out what the kinetic energy of the system is.. Me guess is T_{tot}=T_{rod}+T_{bead}=\dfrac{1}{2}\dot{\theta}^{2}\left(...

Thanks for the quick reply, I am slightly confused as how to in corporate the moment of inertia for the rod. If I use the kinetic energy T=\dfrac{1}{2}M r'^{2} and potential energy U=-mg r \mathrm{cos} \theta there is no difference from the regular pendulum and i don't get the 3/2 factor...

Ach thanks your reference to compound pendulum got me on the right track. It's all about the center of mass: I \theta '' = -M_{cm} g L_{cm} sin \theta \Leftrightarrow \theta '' = -\dfrac{3 g}{2 L} sin (\theta)

Hi all, Homework Statement My problem is a pretty basic one, in a exercise a rigid rod of mass M is rotating around a horizontal pivot point i one end. The rod has the length L. I now need to derive the equation of motion using the Lagrangian formalism. My question is: Can i view the...